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A wire of fixed length is wound on a solenoid of length l and radius R. Its Self Inductance is found to be L. Now, if the same wire is wound on a solenoid of length I2 and radius R2, then self inductance will be


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Solution

Step 1: Given

Originallength=l

Newlength=l2

originalinductance=L

Step 2: To find

We have to find the new inductance.

Step 3: Calculate the new inductance

Use the formula for self-inductance.

The expression for self-inductance is given by,
L=μ0N2Al...(1)
Where, μ0 is the magnetic permeability of free space, N2 is the number of turns per unit length, A is the area of the solenoid, and, l is the length of the solenoid.
The area of a solenoid is given by,

A=πr2
Where, r is the radius of the cross-section of the solenoid.

Substituting this value in equation (1).

L=μ0N2πr2l...(2)

Now, the length of the wire is given by,
length=circumference×numberofturns
l=2πr×N

Since the same wire is taken in both cases, then the circumference and number of turns will be the same for both wires.
L=2πr×N

Rearranging the above equation.
r=L2πN...(3)
Now, substitute equation (3) in equation (2).

L=μ0N2πL2πN2l
L=μ0L24πl
This will be the value of L when the length of the wire is l and radius is r.
For, length l2 and radius r2 value of L will be,
L0=μ0L24πl2

L0=2μ0L24πl
Thus, the self-inductance of the coil will be doubled.

Hence, the new self-inductance will be two times the original self-inductance.


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